Size of neucleus increases from protium to tritium so in `H_(2),D_(2)&T_(2)` area off overlapping also increases in the same order.
Q. `H_(2),D_(2)&T_(2)` show their bond-enthalpies as

Options

A. `H_(2)=D_(2)=T_(2)`
B. `H_(2) gt D_(2) gt T_(2)`
C. `H_(2) lt D_(2) lt T_(2)`
D. `D_(2) lt H_(2) lt T_(2)`

Answer

To solve the question regarding the bond enthalpies of the hydrogen isotopes \( H_2 \), \( D_2 \), and \( T_2 \), we can follow these steps:

### Step 1: Understand the Isotopes
- **Identify the isotopes**:
- \( H_2 \) (Protium) has 1 proton and 0 neutrons.
- \( D_2 \) (Deuterium) has 1 proton and 1 neutron.
- \( T_2 \) (Tritium) has 1 proton and 2 neutrons.

### Step 2: Analyze the Nucleus Size
- **Nucleus Size**: The size of the nucleus increases from \( H_2 \) to \( T_2 \) due to the increasing number of neutrons. This affects the area of overlap between the nuclei when they bond.

### Step 3: Bond Dissociation Energy
- **Bond Dissociation Energy**: The bond dissociation energy is the energy required to break a bond. It varies among the isotopes due to differences in mass and nuclear stability.
- Generally, heavier isotopes have lower bond dissociation energies because the increased mass leads to a greater area of overlap, which results in a stronger bond.

### Step 4: Order of Bond Enthalpies
- **Comparing Bond Enthalpies**:
- For the isotopes, the bond enthalpy order is typically:
- \( T_2 \) (Tritium) < \( D_2 \) (Deuterium) < \( H_2 \) (Protium)
- This means \( H_2 \) has the highest bond enthalpy, followed by \( D_2 \), and \( T_2 \) has the lowest.

### Step 5: Conclusion
- **Final Answer**: The order of bond enthalpies for \( H_2 \), \( D_2 \), and \( T_2 \) is:
\[ \text{Bond Enthalpy Order: } H_2 > D_2 > T_2 \]